# Problem 06 - Variation of Pressure

**Problem**

If the pressure in the tank of oil (sp gr 0.80) is 60 psi, what is the equivalent head: (a) in feet of oil, (b) in feet of water, and (c) in inches of mercury?

**Solution**

Part (a): head in feet of oil

$h_{oil} = \dfrac{p}{\gamma_{oil}}$

$h_{oil} = \dfrac{p}{\gamma_{oil}}$

$h_{oil} = \dfrac{p}{s_{oil} \, \gamma_{water}}$

$h_{oil} = \dfrac{60(12^2)}{0.80(62.4)}$

$h_{oil} = 173.08 \, \text{ft}$ *answer*

Part (b) head in feet of water

$h_{water} = s_{oil} \, h_{oil}$

$h_{water} = 0.80(173.08)$

$h_{water} = 138.46 \, \text{ft}$ *answer*

Part (c) head in inches of mercury

$h_{mercury} = \dfrac{s_{oil}}{s_{mercury}}h_{oil}$

$h_{mercury} = \dfrac{0.80}{13.6}(173.08)$

$h_{mercury} = 10.18 \, \text{ft}$

$h_{mercury} = 122.17 \, \text{in}$ *answer*

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